jimbobjames said:
OK, here is what I understand so far: If you let go of the apple at a point P, we can find a local (free-fall) reference frame there such that the metric (at that point) is that of special relativity and the first derivative of the metric wrt each of the 4 coordinates is zero. (Einstein would say if you fall then there is no gravity, no force). Local experiments over a very short time in that frame would give the same results as those of SR.
By looking at the coordinate transformation from another frame (say as an observer on the surface of the earth) we arrive at the geodesic equation. The metric connections (Christoffel symbols) in that equation depend only on the metric and its derivatives and when they are non-zero we are dealing with a curved manifold.
Overall this is a very good summary. Note that for example, an observer in an accelerating rocketship has non-zero Christoffel symbols (which can be derived from the metric and its first derivative) but a zero Riemann tensor (which can be derived from the metric and its first and second derivatives). For some purposes it is convenient to define a manifold as curved only when the Riemann is non-zero, and flat when the Riemann is zero. With such a definition, an accelerated observer in flat space-time can say that he has non-zero Christoffel symbols and a "flat space-time" (i.e. zero Riemann tensor). So by this defintion, non-zero Christoffel symbols don't imply non-flatness.
In other words, switching to an accelerated frame makes the Christoffel symbols non-zero and the associated frame non-inertial, but by some common defintions of "curvature" (i.e. the nonzero Riemann tensor definition of curvature) it may not mean that space-time becomes curved.
In some other cases, though, writers do talk about the manifold associated with an accelerating observer as if it were curved even though the Riemann is zero. For instance, many textbooks say that gravitational redshift is evidence of "curvature". But this means that the authors of such textbooks must be adopting the "nonzero Christoffel" definition of curvature, because an accelerating observer has a zero Riemann.
In the weak field of the Earth it turns out that only the 0,0 component of the metric differs from that of flat space and GR reduces to Newtonian gravity (due to the curvature of the time part of spacetime).
Yes.
And so, armed with the equivalence principle and some basic calculus on manifolds, we get a lot of results quickly. The rest of Einsteins field equations has to do with showing how energy and mass effect the metric and its (second) derivatives. So you start with the Stress-Energy tensor, work out the metric and then work out the paths.
Maybe my problem lies in thinking there is something locally curved at the apple (the local spacetime) telling it how to move, when in reality it only appears to be so because I am standing on an accelerated frame? - ie from my perspective as an accelerated observer I see (and calculate) the apple's local spacetime to be curved, whereas it is actually flat. I am getting closer now?
Spacetime at the apple is flat, the curvature arises only from my perspective as an accelerated observer?.
There is much confusion in the literature, especially the popularized literature, about what the term "curvature" really means as I remarked earlier. The technical point is that an observer following a geodesic in a Minkowski space-time has locally zero Christoffel symbols, while an accelerating observer in the identical Minkowski space-time has non-zero Christoffel symbols. If one uses the Riemann definition of curvature, however, one would describe the Minkowski space-time as being flat, independent of the observer, so that the space-time is flat both for the accelerating and non-accelerating observer.
Using the non-zero Christoffel symbol definition of curvature would mean that whether or not space-time was curved depended on the observer. This is a rather confusing usage, but some textbooks do it anyway, especially when they talk about the equivalence principle.
Mostly, what I can say is "be aware of the issue".
Pervect, is it your belief (understanding) that there is in fact nothing physical in contact with the apple relaying to it the local curvature? Can you think of an experiment to check if there is anything physical in direct contact with it? (I can't, except maybe beaming something of extremely high energy at that space and checking if whatever is there makes itself known to us?).
Cheers.
With my GR hat on, I would say that there is only space-time. When you start to get into quantum mechanics, though, "empty" space starts to become full of virtual particles, so with a quantum hat on one would probably not take the same perspective about the "emptiness" of the vacuum.
It is also a mystery why cosmological observations suggest that space-time may have a small "cosmological constant" which could be interpreted as "not being empty".
One can separate this issue out by calling the cosmological constant "dark energy". So empty space may possibly still contain this "dark energy". However, the origins of this "dark energy" aren't really well understood, and we have only some cosmological arguments at this point to support its existence. (Though most people do seem to be convinced that the cosmology here is good.)